Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#define ll long long
using namespace std;
int n;
double a;
long double dp[3005][3005];
int main(){
dp[0][0]=1.0;
cin>>n;
for(int i=1;n>=i;i++)
{
	cin>>a;
	for(int j=0;n>=j;j++)
	{
		dp[i][j]=dp[i-1][j]*(1-a);
		if(j)
			dp[i][j]+=dp[i-1][j-1]*a;
	}
}
long double out=0;
int s=n/2+1;
for(int i=s;i<=n;i++)
	out+=dp[n][i];
cout<<fixed<<setprecision(11)<<out;
return 0;
}